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24 Apr 2014, 18:52

Hi All,

I'm getting a little flustered when I try to distinguish between Combination and Permutation. I know the fundamental formulas and what they are *trying* to say -- combination = unordered collection, permutation = ordered collection. Despite *thinking* that I know that means, I have, without a doubt, used them interchangeably.

Is there a different approach to help me distinguish between the two. Bunuel had a great suggestion of trying to encompass them into two sections:

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I'm getting a little flustered when I try to distinguish between Combination and Permutation. I know the fundamental formulas and what they are *trying* to say -- combination = unordered collection, permutation = ordered collection. Despite *thinking* that I know that means, I have, without a doubt, used them interchangeably.

Is there a different approach to help me distinguish between the two. Bunuel had a great suggestion of trying to encompass them into two sections:

Show Tags

24 Apr 2014, 21:50

VeritasPrepKarishma wrote:

russ9 wrote:

Hi All,

I'm getting a little flustered when I try to distinguish between Combination and Permutation. I know the fundamental formulas and what they are *trying* to say -- combination = unordered collection, permutation = ordered collection. Despite *thinking* that I know that means, I have, without a doubt, used them interchangeably.

Is there a different approach to help me distinguish between the two. Bunuel had a great suggestion of trying to encompass them into two sections:

Thanks for sharing the link -- I found it very informative. I feel like I understand it a *little* better and I know exactly what to do once I figured out if it was a permutation vs. a combination problem. My challenge still comes in in making that differentiating.

The example you had about 5 seats and 3 people, that required us to "arrange" in a sense but a very similar problem where we had to choose 3 out of 5, we had to "unarrange". I guess my biggest disconnect is what was the trigger that we had to arrange? Was it because they had given us seats?

Is it safe to say that when the GMAT wants us to PUT something/something IN/ON something -- that usually means that it's an arrange problem, hence a permutation problem?

A random question that I just thought might be relevant? I'm going to paraphrase here but it went something like, "There are 3 red marbles and 7 blue. We pick 3, what is the probability of picking at least 2 blue". I understand that we are adding a probability scenario here but if we just talk about the "favorable outcomes/numerator" and talk about the different combinations, why is it that we have to "arrange" BBR, BRB, RBB when we use the probability equation -- (7/10)(6/9)(3/8) + (7/10)(3/9)(2/8) + ... BUT if we use a Combination/Permutation equation, we don't have to arrange, meaning, we can solve by doing (7C2)(3c1) but we don't need to account for if it's BBR, BRB etc.

Sorry for the long winded question but I hope this helps clarify the topic for other users as well.

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The example you had about 5 seats and 3 people, that required us to "arrange" in a sense but a very similar problem where we had to choose 3 out of 5, we had to "unarrange". I guess my biggest disconnect is what was the trigger that we had to arrange? Was it because they had given us seats?

After selection, if you are going to assign them a place/number/position, you need permutation. If you are just going to carry on with a bunch, this is combination. 5 friends and 3 seats example needed you to select 5 friends and then make them sit on 3 distinct seats - permutationWhen you had to select 3 friends out of 5 to take them to a vacation, you just need a bunch. You are not allotting them any positions - combination

russ9 wrote:

Is it safe to say that when the GMAT wants us to PUT something/something IN/ON something -- that usually means that it's an arrange problem, hence a permutation problem?

A random question that I just thought might be relevant? I'm going to paraphrase here but it went something like, "There are 3 red marbles and 7 blue. We pick 3, what is the probability of picking at least 2 blue". I understand that we are adding a probability scenario here but if we just talk about the "favorable outcomes/numerator" and talk about the different combinations, why is it that we have to "arrange" BBR, BRB, RBB when we use the probability equation -- (7/10)(6/9)(3/8) + (7/10)(3/9)(2/8) + ... BUT if we use a Combination/Permutation equation, we don't have to arrange, meaning, we can solve by doing (7C2)(3c1) but we don't need to account for if it's BBR, BRB etc.

Sorry for the long winded question but I hope this helps clarify the topic for other users as well.

Thanks a ton,Russ

Very hard to generalize the "IN/ON" statement but yeah, I would guess that would be true in most cases.

You consider all arrangements such as BBR/BRB etc because these are the numerous ways in which you can pick marbles.

I am assuming that you meant to write "the probability of picking 2 Blue and 1 Red marbles".In the combination approach, you ignore arrangements because it is a probability question. You CAN arrange in both favorable cases as well as total cases. The ratio of Favorable/Total will stay the same. To arrange the 3 balls you will multiply the numerator by 3!. To arrange the 3 balls in total cases, you will multiply the denominator by 3! too. The 3!s get canceled. That is why you can ignore the arrangements.
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18 May 2014, 10:15

Going with the discussion, could someone explain why in one of the questions order is necessary and in the other it is not? My initial instinct was that order is not necessary for either but I feel I am missing something.

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

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Going with the discussion, could someone explain why in one of the questions order is necessary and in the other it is not? My initial instinct was that order is not necessary for either but I feel I am missing something.

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

In the first one, you do not have to consider the order in which they buy the cars. They select 3 of the 4 colors and then for each color, they just select the model.

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Going with the discussion, could someone explain why in one of the questions order is necessary and in the other it is not? My initial instinct was that order is not necessary for either but I feel I am missing something.

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

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